Exponential population decay word problem

In summary, the conversation discusses a problem where the population of an endangered species is reduced by 25% each year. The formula P(t) = Cekt is used to find the value of k, but the correct value is not initially determined. After some calculations, the correct value is found to be 0.75. The conversation also discusses converting this value into years, which is estimated to be between 2 and 3 years, closer to 2 years. Finally, another method using the formula P0(0.75)t is suggested to solve the problem, resulting in a final answer of 2.1850811 years.
  • #1
Jacobpm64
239
0
Suppose that in any given year, the population of a certain endangered species is reduced by 25%. If the population is now 7500, in how many years will the population be 4000?
Well, I know that the formula for this is P(t) = Cekt.
I just can't figure out the value of k.
i did 7500 x .25 = 1875... 7500 - 1875 = 5625...
so i set up 7500x = 5625 and got .75.
so, essentially.. the population is growing at .75 per year.
so i used .75 as k and:
4000 = 7500e^(.75t)
4000/7500 = e^(.75t)
8/15 = e^(.75t)
ln(8/15) = .75t lne
ln(8/15) = .75t
t = [ln(8/15)] / .75
t = -.8381448792 years
This can't be correct because time isn't negative, and if you do it logically, it would be around 2. something years:
7500 x .75 = 5625 ----> 1 year
5625 x .75 = 4218.75 ----> 2 yrs
4218.75 x .75 = 3164.0625 ----> 3 yrs
So it's somewhere between 2 and 3 years... closer to 2 years.
and if i set up 4218.75x = 4000
x = .9481481481
so i know after 2 years.. i can multiply by .9481481481 and i get 4000..
but i don't know how to convert that into years.. since that is a percentage.
BLEH.. brain is about to explode.. don't know what to do.. lol :bugeye:
 
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  • #2
Should K be -.25? The time comes out to 2.514 then
 
  • #3
There is no reason at all to think that k= 0.75! Don't just plug numbers in at random- think!
Each year, the population decreases by 0.25 and so must be 0.75 times the previous years population. That is, if this year's (t= 1) population is P, next year's is 0.75P: 0.75P= Pek or ek= 0.75. Solve that for k.

Personally, I wouldn't use "ekt" at all. The population is multiplied by 0.75 each year so its population after t years would be P0(0.75)t. You want to solve
7500(0.75)t= 4000.
 
  • #4
got it.. 2.1850811 years
 

What is exponential population decay?

Exponential population decay is a mathematical concept that describes the decrease in a population over time, assuming that the rate of decrease is proportional to the current population size.

What is the formula for exponential population decay?

The formula for exponential population decay is P(t) = P0e^(-rt), where P(t) is the population at time t, P0 is the initial population, e is the base of the natural logarithm, and r is the decay rate.

How do you solve a word problem involving exponential population decay?

To solve a word problem involving exponential population decay, you need to identify the given information, such as the initial population, decay rate, and time period. Then, you can plug these values into the formula P(t) = P0e^(-rt) to find the population at a specific time.

What are some real-life examples of exponential population decay?

Some real-life examples of exponential population decay include the spread of a contagious disease, the decay of radioactive materials, and the decline of a species due to environmental factors.

How does exponential population decay differ from linear population decay?

Exponential population decay differs from linear population decay in that the rate of decrease is constant in linear decay, while it is proportional to the current population size in exponential decay. As a result, exponential decay results in a rapid decrease in population, while linear decay results in a gradual decrease.

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